Chris Malone QFT: HW # 8 October 25, 2006 IIIA3.1 Consider the action Z S[φ] =
dtd
D−1
1 ˙2 x − φ + V (φ) 2
for potential V (φ) (a function, not a functional). a Find the field equations. b Assume V (φ) = λφn for some positive integer n and constant, dimensionless λ, in units ~ = c = 1. Use dimensional analysis to relate n and D (of course, also a positive integer), and list all paired possibilities of (n, D). a To find the field equations, we simply need to set the variation of the action to zero: Z 1 δS[φ] = 0 = δ dtdD−1 x − φ˙ 2 + V (φ) 2 Z i h ˙ φ˙ + δV (φ) = dtdD−1 x −φδ Z ∂V (φ) D−1 ¨ = dtd x φδφ + δφ ∂φ Z ∂V (φ) D−1 ¨ = dtd x δφ φ + ∂φ δS[φ] ∂V (φ) = φ¨ + =0 ∴ δφ ∂φ ∂V (φ) ⇒ φ¨ = − ∂φ b In the units where ~ and c are dimensionless, our action is dimensionless. Therefore the integral Z dtdD−1 x φn is also dimensionless. 1
IIIA4.1 Let’s consider the semiclassical interpretation of a charged particle as described by a complex scalar field ψ, with Lagrangian L=
1 |∇ψ|2 + m2 |ψ|2 2
(1)
a Use the semiclassical expansion in ~ defined by ∇ → ~∂ + iqA,
ψ→
√
ρe−iS/~
Find the Lagrangian in terms of ρ and S (and the background field A), order-by-order in ~ (in this case, just ~0 and ~2 ). b Take the semiclassical limit by dropping the ~2 term in L, to find L→ρ
i 1h (−∂S + qA)2 + m2 2
Vary with respect to S and ρ to find the equations of motion. Defining p ≡ −∂S show that these field equations can be interpreted as the mass-shell condition and current conservation. Show that A couples to this current by varying L with respect to A.
a Using this replacement, we can rewrite (1) as: 2 1 √ −iS/~ 2 2 √ −iS/~ L = ) + m ρe (~∂ + iqA)( ρe 2 1 √ −iS/~ 2 2 = ) + m ρ (~∂ + iqA)( ρe 2 ! 2 1 ~∂ρ √ = √ + i ρ(qA − ∂S) e−iS/~ + m2 ρ 2 2 ρ 1 ~2 2 2 2 = (∂ρ) + ρ(qA − ∂S) + m ρ 2 4ρ ! ρ 1 ∂ρ 2 2 L = ~ + (qA − ∂S)2 + m2 ~0 . 2 4 ρ
2
(2)
b Now, if we drop the terms in (2) of order ~2 clearly we have L=
ρ (−∂S + qA)2 + m2 . 2
(3)
To get the equations of motion, we need to solve the Euler-Lagrange equations: ∂L ∂L ∂L ∂L −∂ = 0, −∂ = 0. ∂ρ ∂(∂ρ) ∂S ∂(∂S) We see that (3) has no ∂ρ term and also no S term. Therefore we can neglect the partial derivatives with respect to these variables. The other terms become: ∂L ∂ρ −∂
∂L ∂(∂S)
1 (−∂S + qA)2 + m2 = 0 2 ⇒ (−∂S + qA)2 + m2 = 0 =
=
(4)
∂ (ρ [−∂S + qA]) = 0
⇒ ∂ (−ρ∂S + qρA) = 0.
(5)
Now, if we make the substitution for −∂S as stated in the problem, we see (4) becomes (6) (p + qA)2 + m2 = 0 which is exactly the mass shell condition, where the total momentum includes terms from the gauge field. Similarly, plugging into (5) gives ∂ (ρp + qρA) = ∂J = 0
(7)
or that the current (again including terms due to the gauge field) is a conserved quantity. For the A coupling we have ∂L = qρ (p + qA) = qJ = 0. ∂A
(8)
IIIA4.3 By plugging in the appropriate expressions in terms of Aa (and repeatedly integrating by parts), show that all of the above expressions for the electromagnetism action can be written as Z 1 SA = − dx 2 [A · 2A + (∂ · A)2 ] 4e
3
If we start from the expression in the book: Z 1 SA = dx 2 F ab Fab + Aa Ja 8e where J b = 2e12 ∂a F ab is the current density and expand the field tensor we have Z Z 1 1 a b b a SA = dx 2 ∂ A − ∂ A (∂a Ab − ∂b Aa ) + 2 dx Aa ∂ α (∂α Aa − ∂a Aα ) 8e 2e Z 1 a b = dx 2 ∂ A ∂a Ab − ∂ a Ab ∂b Aa − ∂ b Aa ∂a Ab + ∂ b Aa ∂b Ab 8e Z 1 + 2 dx Aa ∂ α (∂α Aa − ∂a Aα ) 2e Z h i 1 a b a b a α = dx (∂ A )(∂ A ) − (∂ A )(∂ A ) + 2 {A ∂ (∂ A − ∂ A )} α a a α a a b b 4e2 Z i h 1 a b a α = dx −A · 2A − (∂ A )(∂ A ) + 2A · 2A − 2A ∂ ∂ A a a α b 4e2 Z i h 1 a b a α = dx A · 2A − (∂ A )(∂ A ) − 2A ∂ ∂ A a a α b 4e2
4
